A box of 1 L capacity is divided into two equal compartments by a thin partition which are filled with 2g `H_(2)` and 16 g`CH_(4)` respectively. The pressure in each compartment is reorded as P atm. The total pressure when partition is removed will be:

Options

A. P
B. 2P
C. `P//2`
D. `P//4`

Explanation

To solve the problem step by step, we will use the ideal gas law and the concept of pressure in a gas mixture.

### Step 1: Understand the Setup
We have a box of 1 L capacity divided into two compartments. One compartment contains 2 g of \( H_2 \) and the other contains 16 g of \( CH_4 \). The pressure in each compartment is recorded as \( P \) atm.

### Step 2: Calculate the Number of Moles
We need to calculate the number of moles of each gas using the formula:
\[ n = \frac{\text{mass}}{\text{molar mass}} \]

- For \( H_2 \):
- Molar mass of \( H_2 = 2 \, \text{g/mol} \)
- Number of moles of \( H_2 \):
\[ n_1 = \frac{2 \, \text{g}}{2 \, \text{g/mol}} = 1 \, \text{mol} \]

- For \( CH_4 \):
- Molar mass of \( CH_4 = 16 \, \text{g/mol} \)
- Number of moles of \( CH_4 \):
\[ n_2 = \frac{16 \, \text{g}}{16 \, \text{g/mol}} = 1 \, \text{mol} \]

### Step 3: Initial Conditions
Each compartment has a volume of \( \frac{1}{2} \, \text{L} \) and the pressure in each is \( P \) atm. We can use the ideal gas law \( PV = nRT \) to express the temperature in terms of pressure.

For one compartment (either \( H_2 \) or \( CH_4 \)):
\[ P \cdot \frac{1}{2} = nRT \implies T = \frac{P \cdot \frac{1}{2}}{nR} \]

Since \( n = 1 \, \text{mol} \):
\[ T = \frac{P \cdot \frac{1}{2}}{1 \cdot R} = \frac{P}{2R} \]

### Step 4: Total Moles After Removing the Partition
When the partition is removed, the total volume becomes 1 L and the total number of moles is:
\[ n_{\text{total}} = n_1 + n_2 = 1 + 1 = 2 \, \text{mol} \]

### Step 5: Calculate the Total Pressure
Using the ideal gas law again for the total system:
\[ P_{\text{total}} \cdot V_{\text{total}} = n_{\text{total}} \cdot R \cdot T \]
Substituting \( V_{\text{total}} = 1 \, \text{L} \) and \( T = \frac{P}{2R} \):
\[ P_{\text{total}} \cdot 1 = 2 \cdot R \cdot \frac{P}{2R} \]

This simplifies to:
\[ P_{\text{total}} = P \]

### Conclusion
The total pressure when the partition is removed will be \( P \) atm.

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